Finding a Triple Integral Defined over a Closed Box

Find \(\iiint\limits_{\kern-3ptE}4xyz\,{dV}\) if \(E\) is the closed box \(0\leq x\leq 1\), \(0\leq y\leq 2\), and \(0\leq z\leq 3\) shown in Figure 42.

Solution Since \(f(x,y,z) =4xyz\) is continuous on \(E,\) we use Fubini’s Theorem for triple integrals. \[ \begin{eqnarray*} \displaystyle\iiint\limits_{\kern-3ptE}4xyz\,{\it dV} &=&\int_{0}^{1}\int_{0}^{2}\int_{0}^{3}4xyz\,{\it dz}\,{\it dy}\,{\it dx}=\int_{0}^{1} \int_{0}^{2}4xy\left[ \frac{z^{2}}{2}\right] _{0}^{3}\,{\it dy}\,{\it dx}\\ &=&\int_{0}^{1}\int_{0}^{2}18xy\,{\it dy}\,{\it dx}=\int_{0}^{1}18x\left[ \frac{y^{2}}{2}\right]_{0}^{2}\,{\it dx} \notag \\ &=&\int_{0}^{1}36x\,{\it dx}=\big[ 18x^{2}\big] _{0}^{1}=18 \end{eqnarray*} \]